Chapter 5.3, Problem 7QR

To determine

To find : The horizontal asymptotes of the function’s graph.

Answer to Problem 7QR

The horizontal asymptotes is $y=0$ .

Explanation of Solution

Given information :

The given function is $y=\left(4-x^2\right)e^x$ .

Calculation:

The line $y=b$ is the horizontal asymptotes of the graph of the function $y=f\left(x\right)$ if either

  $\underset{x\to \infty }{\lim }f\left(x\right)=b$ or $\underset{x\to -\infty }{\lim }f\left(x\right)=b$ .

Let $f\left(x\right)=\left(4-x^2\right)e^x$ , then

  $\begin{array}{l}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(4-x^2\right)e^x\\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(4-x^2\right)\times \underset{x\to \infty }{\lim }{e}^x\\ \underset{x\to \infty }{\lim }f\left(x\right)=\left(4-{\infty }^2\right)\times e^{\infty }\\ \underset{x\to \infty }{\lim }f\left(x\right)=-\infty \end{array}$

Now,

  $\begin{array}{l}\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(4-x^2\right)e^x\\ \underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(4-x^2\right)\times \underset{x\to -\infty }{\lim }{e}^x\\ \underset{x\to -\infty }{\lim }f\left(x\right)=\left(4-{\left(-\infty \right)}^2\right)\times e^{-\infty }\\ \underset{x\to -\infty }{\lim }f\left(x\right)=\left(4-{\left(-\infty \right)}^2\right)\times \frac{1}{e^{\infty }}\\ \underset{x\to -\infty }{\lim }f\left(x\right)=0\end{array}$

Since, $\underset{x\to -\infty }{\lim }f\left(x\right)=0$ , therefore the horizontal asymptotes is $y=0$ .

Hence,

The horizontal asymptotes is $y=0$ .